Question: Find the distance between the point ${(3, -3)}$ and the line $\enspace {y = \dfrac{1}{2}x - 2}\thinspace$. ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$ ${1}$ ${2}$ ${3}$ ${4}$ ${5}$ ${6}$ ${7}$ ${8}$ ${9}$ ${\llap{-}2}$ ${\llap{-}3}$ ${\llap{-}4}$ ${\llap{-}5}$ ${\llap{-}6}$ ${\llap{-}7}$ ${\llap{-}8}$ ${\llap{-}9}$
Answer: First, find the equation of the perpendicular line that passes through ${(3, -3)}$ The slope of the blue line is ${\dfrac{1}{2}}$ , and its negative reciprocal is ${-2}$ Thus, the equation of our perpendicular line will be of the form $\enspace {y = -2x + b}\thinspace$ We can plug our point, ${(3, -3)}$ , into this equation to solve for ${b}$ , the y-intercept. $-3 = {-2}(3) + {b}$ $-3 = -6 + {b}$ $-3 + 6 = {b} = 3$ The equation of the perpendicular line is $\enspace {y = -2x + 3}\thinspace$ We can see from the graph (or by setting the equations equal to one another) that the two lines intersect at the point ${(2, -1)}$ . Thus, the distance we're looking for is the distance between the two red points. The distance formula tells us that the distance between two points is equal to: $\sqrt{( x_{1} - x_{2} )^2 + ( y_{1} - y_{2} )^2}$ Plugging in our points ${(3, -3)}$ and ${(2, -1)}$ gives us: $\sqrt{( {3} - {2} )^2 + ( {-3} - {-1} )^2}$ $= \sqrt{( 1 )^2 + ( -2 )^2} = \sqrt{5} $ The distance between the point ${(3, -3)}$ and the line $\thinspace {y = \dfrac{1}{2}x - 2}\enspace$ is $\thinspace\sqrt{5}$.